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6201-16000-MATH-2318 Afeez Amusan & Time Remaining: Quiz: Quiz 2 (1.3, 1.4), Part 1 This Question: 7 pts 11 of 17 (7 complete
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72 4 -7 16 -1 -1 1 -3 | 0 -6 15 -30 { 2 0 3 -6/

R2 + R2+

R4 - R4-1: R

(2 4 0 1 0 -6 lo -4 -7 - 5 15 10 -30 -22

R2 + R

24 -7 16 0 -6 15 -30 0 135 10 -4 10 -22,

R3 + R3+ RE

R4 + R4- R

(2 o o \o 4 — 6 o o -7 15 o o 16 – 30 o -2 /

Rat R

=\begin{pmatrix}2&4&-7&16\\ 0&-6&15&-30\\ 0&0&0&0\\ 0&0&0&1\end{pmatrix}

Ra - R430 R

R_1\:\leftarrow \:R_1-16\cdot \:R_4

(2 4 о –6 To o (o o -7 o\ 15 о o o o 1

R_3\:\leftrightarrow \:R_4

=\begin{pmatrix}2&4&-7&0\\ 0&-6&15&0\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}

R_2\:\leftarrow \:-\frac{1}{6}\cdot \:R_2

72 4 —7 0\ | о 1 - о| | o o o 1 \o o o o/

R_1\:\leftarrow \:R_1-4\cdot \:R_2

=\begin{pmatrix}2&0&3&0\\ 0&1&-\frac{5}{2}&0\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}

R_1\:\leftarrow \frac{1}{2}\cdot \:R_1

=\begin{pmatrix}{\color{Red} 1}&0&\frac{3}{2}&0\\ 0&{\color{Red} 1}&-\frac{5}{2}&0\\ 0&0&0&{\color{Red} 1}\\ 0&0&0&0\end{pmatrix}.............................reduced row echelon form

there is no pivot entry at third column

so columns of matrix A are linearly dependent

so NO we cannot write can each vector in R4 be written as a linear combination of the columns of the matrix A

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do the columns of A span R4 ?

NO because not all b in Rm can be written as a linear combination of the columns of the matrix A

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